The proof will be divided into two parts. In the first part, we will prove the case a = 0. Then, in the second part, we will prove the case a > 0.
- Suppose a = 0. Let $(x_n)$ be a sequence of $(0, \infty)$ such that $x_n \longrightarrow 0$. Given $\varepsilon > 0$, there exists $n_0 \in \mathbb{N}$ such that
$$n > n_0 \ \Rightarrow \ \mid x_n - a \mid \ < \ \varepsilon^s.$$
Follow thats
$$n > n_0 \ \Rightarrow \ \sqrt[s]{x_n} \ = \ \mid \sqrt[s]{x_n} - 0 \mid \ < \ \sqrt[s]{\varepsilon^s} \ = \ \varepsilon. $$
Therefore, $\lim\limits_{x \longrightarrow \ 0} \sqrt[s]{x} = 0$.
- Suppose a > 0. Let $(x_n)$ be a sequence of $[0, \infty) - \{a\}$ such that $x_n \longrightarrow a$. Given $\varepsilon > 0$, there exists $n_0 \in \mathbb{N}$ such that
$$n > n_0 \ \Rightarrow \ \mid x_n - a \mid \ < \ \varepsilon\sqrt[s]{a^{s-1}}.$$
Note that
$$x_n - a = (\sqrt[s]{x_n} - \sqrt[s]{a})(\sqrt[s]{x_{n}^{s - 1}} + \sqrt[s]{x_n^{s - 2}}\sqrt[s]{a} + ... + \sqrt[s]{x_n}\sqrt[s]{a^{s - 2}} + \sqrt[s]{a^{s-1}}).$$
Realize that
$$\sqrt[s]{a^{s-1}} \ \leq \ \sqrt[s]{x_{n}^{s - 1}} + \sqrt[s]{x_n^{s - 2}}\sqrt[s]{a} + ... + \sqrt[s]{x_n}\sqrt[s]{a^{s - 2}} + \sqrt[s]{a^{s-1}} \ \Rightarrow \ \dfrac{1}{\sqrt[s]{x_{n}^{s - 1}} + \sqrt[s]{x_n^{s - 2}}\sqrt[s]{a} + ... + \sqrt[s]{x_n}\sqrt[s]{a^{s - 2}} + \sqrt[s]{a^{s-1}}} \ \leq \ \dfrac{1}{\sqrt[s]{a^{s-1}}}. $$
Then
$$n > n_0 \ \Rightarrow \ \mid x_n - a \mid \ \cdot \ \dfrac{1}{\sqrt[s]{x_{n}^{s - 1}} + \sqrt[s]{x_n^{s - 2}}\sqrt[s]{a} + ... + \sqrt[s]{x_n}\sqrt[s]{a^{s - 2}} + \sqrt[s]{a^{s-1}}} $$
$$ = \dfrac{\mid x_n - a \mid}{\sqrt[s]{x_{n}^{s - 1}} + \sqrt[s]{x_n^{s - 2}}\sqrt[s]{a} + ... + \sqrt[s]{x_n}\sqrt[s]{a^{s - 2}} + \sqrt[s]{a^{s-1}}} \ = \ \mid \sqrt[s]{x_n} - \sqrt[s]{a} \mid \ < \ \dfrac{\varepsilon\sqrt[s]{a^{s-1}}}{\sqrt[s]{a^{s-1}}} = \varepsilon.$$
Therefore, $\lim\limits_{x \longrightarrow \ a} \sqrt[s]{x} = \sqrt[s]{a}$. Then we finish the proof.
It's true?